Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int M = 10050, N = 29;
int d[N + 2], ind[M], b[N + 2], x[M], y[M], dx[N + 2], dy[M];
vector<pair<int, int> > ans, t;
int n, cnt, s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &x[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &y[i]);
  for (int i = 1; i <= n; ++i) {
    ind[i] = -1;
    for (int j = N, s = 0; j >= 0; --j)
      if (x[i] >> j & 1)
        if (d[j]) {
          x[i] ^= x[d[j]];
          s ^= dx[j];
        } else {
          ind[i] = j;
          d[j] = i;
          dx[j] = s | (1 << j);
          break;
        }
  }
  for (int i = 1; i <= n; ++i) {
    s = 0;
    for (int j = N; j >= 0; --j)
      if (y[i] >> j & 1)
        if (d[j]) {
          y[i] ^= x[d[j]];
          s ^= dx[j];
        } else {
          puts("-1");
          return 0;
        }
    if (ind[i] >= 0) {
      b[cnt] = i;
      dy[cnt++] = s;
      continue;
    }
    ans.push_back(make_pair(i, i));
    for (int j = N; j >= 0; --j)
      if (s >> j & 1) ans.push_back(make_pair(i, d[j]));
  }
  for (int i = 0; i < cnt; ++i) {
    int s = dy[i];
    dy[i] = 0;
    for (int j = 0; j < cnt; ++j)
      if (s >> ind[b[j]] & 1) dy[i] |= 1 << j;
  }
  for (int i = 0; i < cnt; ++i) {
    if (!(dy[i] >> i & 1))
      for (int j = i + 1; j < cnt; ++j)
        if (dy[j] >> i & 1) {
          t.push_back(make_pair(b[i], b[j]));
          t.push_back(make_pair(b[j], b[i]));
          t.push_back(make_pair(b[i], b[j]));
          swap(dy[i], dy[j]);
          break;
        }
    if (dy[i] >> i & 1)
      for (int j = i + 1; j < cnt; ++j)
        if (dy[j] >> i & 1) {
          t.push_back(make_pair(b[j], b[i]));
          dy[j] ^= dy[i];
        }
  }
  for (int i = 0; i < cnt; ++i) {
    if (!(dy[i] >> i & 1)) ans.push_back(make_pair(b[i], b[i]));
    for (int j = i + 1; j < cnt; ++j)
      if (dy[i] >> j & 1) ans.push_back(make_pair(b[i], b[j]));
  }
  reverse(t.begin(), t.end());
  for (auto v : t) ans.push_back(v);
  printf("%d\n", ans.size());
  for (auto v : ans) printf("%d %d\n", v.first, v.second);
  return 0;
}